Homework Answers
From the given data,
Molarity of lead nitrate = 0.088 M
Volume of lead nitrate = 6.0 mL
Hence,
Number of moles of lead nitrate = (molarity)(volume in L)
= ( 0.088M)(6/1000L)
= 0.000528 mol.
At intial the total volume of solution = 6.0mL +10mL = 16mL
Concentration of lead nitrate =(moles)÷(volume in L)
= (0.000528mol)÷(16/1000 L)
= 0.033M
Hence,
Concentration of lead nitrate = 0.033 M.
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